问题
My team and I are dealing with many thousands of URLs that have similar segments. Some URLs have one segment ("seg", plural, "segs") in a position of interest to us. Other similar URLs have a different seg in the position of interest to us. We need to sort a dataframe consisting of URLs and associated unique segs in the position of interest, showing the frequency of those unique segs.
Here is a simplified example:
url <- c(1, 3, 1, 4, 2, 3, 1, 3, 3, 3, 3, 2)
seg <- c("a", "c", "a", "d", "b", "c", "a", "x", "x", "y", "c", "b")
df <- data.frame(url,seg)
We are looking for the following:
url freq seg
1 3 a in other words, url #1 appears three times each with a seg = "a",
2 2 b in other words: url #2 appears twice each with a seg = "b",
3 3 c in other words: url #3 appears three times with a seg = "c",
3 2 x two times with a seg = "x", and,
3 1 y once with a seg = "y"
4 1 d etc.
I can get there using a loop and several small steps, but am convinced there is a more elegant way of doing this. Here's my inelegant approach:
Create empty dataframe with num.unique rows and three columns (url, freq, seg)
result <- data.frame(url=0, Freq=0, seg=0)
Determine the unique URLs
unique.df.url <- unique(df$url)
Loop through the dataframe
for (xx in unique.df.url) {
url.seg <- df[which(df$url == unique.df.url[xx]), ] # create a dataframe for each of the unique urls and associated segs
freq.df.url <- data.frame(table(url.seg)) # summarize the frequency distribution of the segs by url
result <- rbind(result,freq.df.url) # append a new data.frame onto the last one
}
Eliminate rows in the dataframe where Frequency = 0
result.freq <- result[which(result$Freq |0), ]
Sort the dataframe by URL
result.order <- result.freq[order(result.freq$url), ]
This yields the desired results, but since it is so inelegant, I am concerned that once we move to scale, the time required will be prohibitive or at least a concern. Any suggestions?
回答1:
In base R you can do this :
aggregate(freq~seg+url,`$<-`(df,freq,1),sum)
# or aggregate(freq~seg+url, data.frame(df,freq=1),sum)
# seg url freq
# 1 a 1 3
# 2 b 2 2
# 3 c 3 3
# 4 x 3 2
# 5 y 3 1
# 6 d 4 1
The trick with $<- is just to add a column freq of value 1 everywhere, without changing your source table.
Another possibility:
subset(as.data.frame(table(df[2:1])),Freq!=0)
# seg url Freq
# 1 a 1 3
# 8 b 2 2
# 15 c 3 3
# 17 x 3 2
# 18 y 3 1
# 22 d 4 1
Here I use [2:1] to switch the order of columns so table orders the results in the required way.
回答2:
url <- c(1, 3, 1, 4, 2, 3, 1, 3, 3, 3, 3, 2)
seg <- c("a", "c", "a", "d", "b", "c", "a", "x", "x", "y", "c", "b")
df <- data.frame(url,seg)
library(dplyr)
df %>% count(url, seg) %>% arrange(url, desc(n))
# # A tibble: 6 x 3
# url seg n
# <dbl> <fct> <int>
# 1 1 a 3
# 2 2 b 2
# 3 3 c 3
# 4 3 x 2
# 5 3 y 1
# 6 4 d 1
回答3:
Would the following code be better for you?
library(dplyr)
df %>% group_by(url, seg) %>% summarise(n())
回答4:
Or paste & tapply:
url <- c(1, 3, 1, 4, 2, 3, 1, 3, 3, 3, 3, 2)
seg <- c("a", "c", "a", "d", "b", "c", "a", "x", "x", "y", "c", "b")
df <- data.frame(url,seg)
want <- tapply(url, INDEX = paste(url, seg, sep = "_"), length)
want <- data.frame(do.call(rbind, strsplit(names(want), "_")), want)
colnames(want) <- c("url", "seg", "freq")
want <- want[order(want$url, -want$freq), ]
rownames(want) <- NULL # needed?
want <- want[ , c("url", "freq", "seg")] # needed?
want
回答5:
An option can be to use table and tidyr::gather to get data in format needed by OP:
library(tidyverse)
table(df) %>% as.data.frame() %>%
filter(Freq > 0 ) %>%
arrange(url, desc(Freq))
# url seg Freq
# 1 1 a 3
# 2 2 b 2
# 3 3 c 3
# 4 3 x 2
# 5 3 y 1
# 6 4 d 1
OR
df %>% group_by(url, seg) %>%
summarise(freq = n()) %>%
arrange(url, desc(freq))
# # A tibble: 6 x 3
# # Groups: url [4]
# url seg freq
# <dbl> <fctr> <int>
# 1 1.00 a 3
# 2 2.00 b 2
# 3 3.00 c 3
# 4 3.00 x 2
# 5 3.00 y 1
# 6 4.00 d 1
来源：https://stackoverflow.com/questions/50492862/ordering-a-dataframe-by-its-subsegments